\(n_{CH_4}=\frac{10,2}{22,4}=0,45\left(mol\right)\)
a. PTHH: \(CH_4+2O_2-to->CO_2+2H_2O\)
Theo PT: \(n_{O_2}=0,45.2=0,9\left(mol\right)\)
=> \(V_{O_2\left(cần-dùng\right)}=0,9.22,4=20,16\left(l\right)\)
nCH4= 10.2/22.4=51/112 mol
CH4 + 2O2 -to-> CO2 + 2H2O
51/112___51/56
VO2= 51/56*22.4=20.4l